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sicp exercise 13

Name: Anonymous 2009-07-04 13:30

sicp's exercise no. 13
what do i need to solve this? can i do it with mathematical induction?
i saw that last semester, but didn't quite get it. i'll review it if it's necessary.

Name: Anonymous 2009-07-04 13:34

what about sicp is excercise no 13?

Name: Anonymous 2009-07-04 13:50

>>2
FUCK YOU

Name: Anonymous 2009-07-04 13:58

>>3
What about FUCK?

Name: Anonymous 2009-07-04 14:15

>>4
What about FUCKQuestionMark?

Name: Anonymous 2009-07-04 14:48

>>2
did i get that wrong? i'm sorry, english isn't my first language. although i think you understood what i meant.

Name: Anonymous 2009-07-04 14:57

>>6
YHBT

Name: Anonymous 2009-07-04 15:29

>>7
YHBT

Name: Anonymous 2009-07-04 15:51

>>8

you have been trolled

Name: Anonymous 2009-07-04 15:57

/tail recursion

Name: Anonymous 2009-07-04 16:51

>>10
sure. i'm going to do a mathematical proof with tail recursion. how does that make sense?
btw, i meant exercise 1.13

Name: Anonymous 2009-07-04 21:13

φ = (1 + √5) / 2
ψ = (1 - √5) / 2

φ · φ = ((1 + √5) / 2) · ((1 + √5) / 2)
      = ((1 + √5) · (1 + √5)) / (2 · 2)
      = (1 · 1 + √5 · √5 + 1 · √5 + √5 · 1) / 4
      = (1 + 5 + 2 · √5) / 4
      = (2 + 2 · √5) / 4 + 4 / 4
      = (1 + √5) / 2 + 1
      = φ + 1

ψ · ψ = ((1 - √5) / 2) · ((1 - √5) / 2)
      = ((1 - √5) · (1 - √5)) / (2 · 2)
      = (1 · 1 + -√5 · -√5 + 1 · -√5 + -√5 · 1) / 4
      = (1 + 5 + -2 · √5) / 4
      = (2 + -2 · √5) / 4 + 4 / 4
      = (1 - √5) / 2 + 1
      = ψ + 1

Fib(0) = (φ0 - ψ0) / √5
       = 0

Fib(1) = (φ1 - ψ1) / √5
       = (((1 + √5) / 2) - ((1 - √5) / 2)) / √5
       = ((1 + √5) - (1 - √5)) / (2 · √5)
       = (2 · √5) / (2 · √5)
       = 1

Fib(n+2) = Fib(n) + Fib(n+1)
         = (φn - ψn) / √5 + (φn+1 - ψn+1) / √5
         = (φn - ψn + φn+1 - ψn+1) / √5
         = (φn - ψn + φn · φ - ψn · ψ) / √5
         = (φn · (φ + 1) - ψn · (ψ + 1)) / √5
         = (φn · φ · φ - ψn · ψ · ψ) / √5
         = (φn+2 - ψn+2) / √5

|Fib(n) - (φn / √5)| = |ψn / √5|

n / √5| < ½

|Fib(n) - (φn / √5)| < ½

Name: Anonymous 2009-07-04 22:42

>>12

lrn2 optimize effort.

1) Download/Open Maple or some other math program that does symbolic manipulation.

2) Verify that (φ^0 - ψ^0) / √5 = 0, (φ^1 - ψ^1) / √5 = 1, and (φ^n - ψ^n) / √5 + (φ^(n+1) - ψ^(n+1)) / √5.  (http://i39.tinypic.com/350nixi.jpg)

3) Since Fib(n) = (φ^n - ψ^n) / √5 and φ^n / √5 differ by a quantity less than .5, Fib(n) is the nearest integer.

Name: Anonymous 2009-07-04 22:57

>>13
Maple
Now you have two problems.

Name: Anonymous 2009-07-04 23:01

>>6
You have been trolled.
Also, use capitals. You look retarded when you omit them.

Name: Anonymous 2009-07-04 23:11

>>14
I use Mathematica whenever possible, but Maple has the WYSIWYG input scheme, which offsets it's overall shittiness for small problems sometimes.

Name: Anonymous 2009-07-05 0:19

This thread is threadstopped. You can't reply anymore.

Name: Anonymous 2009-07-05 1:15

>>15
but ȝe lōkes frencisc ƕen ȝe brǒukes boðe capitāles a minuscules.
brǒuke ōn and not ðē tōðer.

Name: Anonymous 2009-07-05 1:40

Whoa! I've never seen a “ƕ” before.

Name: Anonymous 2009-07-05 1:50

ƕƕƕƕƕƕƕƕƕƕƕƕƕƕƕƕƕƕƕƕƕƕ

Name: Anonymous 2009-07-21 3:12

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Name: Anonymous 2010-12-06 9:55

Back to /b/, ``GNAA Faggot''

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